Re: [微積] Test for convergence

看板Math作者Sfly (topos)時間14年前 (2012/02/09 18:53)推噓4(4推 0噓 1→)

留言5則, 3人參與討論串2/2 (看更多)

※ 引述《tasukuchiyan (Tasuku)》之銘言： : 1.Σ(1．4．．．(3n-2)/(3．6．．．3n))^2 : 2.Σ(1．3．．．(2n-1)/(2．4．．．2n))^2 : 請問這兩題要怎麼驗證它是否收斂？有任何想法嗎？謝謝。 By using taylor expansion of log(1+x), one can see easily that (1．4．．．(3n-2)/(3．6．．．3n))^2 = O(1/n^(4/3)) and (1．3．．．(2n-1)/(2．4．．．2n))^2 = O(1/n). So, the first series converges and the 2nd diverges. For basic approach: 1 4 (3n-2) 1.let a= -- x --- x ....x-------- 3 6 3n 2 5 (3n-1) < b= -- x --- x ....x-------- 4 7 3n+1 3 6 3n < c= -- x --- x ....x-----. 5 8 3n+2 hence abc= 2/(3n+1)(3n+2), so a^2< (abc)^(2/3) = O(1/n^(4/3)). Thus, series in Q1 converges. 3*5*.......*(2n-1) 4*6*........*(2n) 2.let a= ----------------------- > b= -----------------------. 2*4*.......*(2n-2)*2n 3*5*........*(2n-1)*2n Hence a^2> ab=1/4n. Thus, series 2 diverges. -- 哪一首中文歌的歌詞有 Pasadena? -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 76.94.119.209

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bineapple

02/09 22:53, , 1^F

02/09 22:53, 1^F

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WINDHEAD

02/10 01:24, , 2^F

02/10 01:24, 2^F

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tasukuchiyan

02/10 12:32, , 3^F

02/10 12:32, 3^F